题目：Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
分析：
--->题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。
--->思路：遍历一遍链表，把小于x的都挂到part1后，把大于等于x的都放到part2后，最后再把大于等于的链表挂到小于链表的后面就可以了。
代码：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode *part1 = new ListNode(0), *part2 = new ListNode(0), *node1, *node2;
node1 = part1;
node2 = part2;
while (head) {
if (head->val < x) {
node1->next = head;
node1 = node1->next;
} else {
node2->next = head;
node2 = node2->next;
}
head = head->next;
}
node2->next = NULL;
node1->next = part2->next;
return part1->next;
}
};
原文：http://www.cnblogs.com/carsonzhu/p/5143694.html